CK-12-Chemistry Intermediate

http://www.ck12.org Chapter 14. The Behavior of Gases

Graham’s Law

When a person opens a bottle of perfume in one corner of a large room, it doesn’t take very long for the scent
to spread throughout the entire room. Molecules of the perfume evaporate and the vapor spreads out to fill the
entire space.Diffusionis the tendency of molecules to move from an area of high concentration to an area of low
concentration until the concentration is uniform. While gases diffuse rather quickly, liquids diffuse much more
slowly (Figure14.15). Solids essentially do not diffuse.

FIGURE 14.15

Two liquids, yellow and green, diffuse into one another.

Another related process is effusion. Effusionis the process of a confined gas escaping through a tiny hole in its
container. Effusion can be observed by the fact that a helium-filled balloon will stop floating and sink to the floor
after a day or so. This is because the helium gas effuses through tiny pores in the balloon. Both diffusion and
effusion are related to the speed at which various gas molecules move. Gases that have a lower molar mass effuse
and diffuse at a faster rate than gases that have a higher molar mass.

Scottish chemist, Thomas Graham (1805-1869), studied the rates of effusion and diffusion for various gases.Gra-
ham’s Lawstates that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the
molar mass of the gas. Graham’s Law can be understood by comparing two gases (A and B) at the same temperature,
meaning the gases have the same average kinetic energy. The kinetic energy of a moving object is given by the
equation KE=^12 mv^2 , wheremis mass andvis velocity. Setting the kinetic energies of the two gases equal to one
another gives:

1

2 mAv

2

A=

1

2 mBv

2

B

The equation can be rearranged to solve for the ratio of the velocity of gas A to the velocity of gas B (vA/vB).

v^2 A

v^2 B

=

mB

mA

which becomes

vA

vB

=A

For the purposes of comparing the rates of effusion or diffusion for two gases at the same temperature, the molar
masses of each gas can be used in the equation form.

Sample Problem 14.10: Graham’s Law

Calculate the ratio of diffusion rates for ammonia gas (NH 3 ) and hydrogen chloride (HCl) at the same temperature
and pressure.